Problem: $a(n) = \dfrac{3}{2} \left(-2\right)^{n - 1}$ What is the $3^\text{rd}$ term in the sequence?
Explanation: This is an explicit formula. All we have to do is plug $n=3$ in the formula to find the $3^\text{rd}$ term. $\begin{aligned} a({3}) &=\dfrac32(-2)^{{3} - 1} \\\\ &= 6 \end{aligned}$ The $3^\text{rd}$ term is $6$.